Basic Poultry Genetics: A Quick Read From Netherlands

As breeder-cockfighters, we select for winning performance in the cockpits. Then we begin to wonder how to best breed our chickens to look the same and to fight the same. Without the knowledge of basic chicken genetics, one can breed using common sense but it will be a longer process of experimentation and observation.

With basic genetics knowledge,

We can easily determine what possible old breeds were used to develop a new breed

We can easily determine the possible physical phenotype of the offsprings from a pair mating

We can easily calculate the possible genotype of the offsprings from a pair mating

We can easily verify the true genetic make up of a newly acquired gamecock chicken

I always test the purity of newly acquired chickens

based both on their breeder-cockfighter’s description of the breed

and on my own research including but not limited to actual pit winners’ characteristics.

General Introduction into Poultry Genetics:

“Sexual dimorphism” includes differing colour/pattern forms between the two sexes of a species, i.e. not restricted to physical body traits. The gender colour/pattern differences in the wild type – Red Jungle Fowl (black breasted red roosters, salmon breasted hens, etc) are two alternate colour/pattern forms, the particular form expressed dependent on gender.

P1

Parent 1 (in first cross)

F1 F2……

First Filial, Second Filial (1st Generation, etc)

BC1…. etc:

Back Cross 1 (cross F1 back to Parent, etc)

Gene

A unit of hereditary , a section of DNA found on a chromosome that codes for a particular protein

Locus (Loci – plural)

The location of an Allele on the Chromosome

Allele

One of two alternate forms of a gene that has the same locus on homologous chromosomes

Chromosome

A threadlike body in the cell nucleus that carries the genes in a linear order

Homozygous

Where alleles of a locus on homologous chromosomes are the same

Heterozygous

Where alleles of a locus on homologous chromosomes are different

Hemizygous

A genetic locus present in one copy only. Of females, in reference to the only one allele at each locus on the single Z chromosome (only one allele possible, as only one Z chromosome in females, therefore not heterozygous or homozygous as no chromosome pair).

Linkage Map Unit, (centiMorgan)

1 map unit, or 1 centiMorgan (cM) is equal to 1% recombination

Inheritance Mode Types:

Dominant:

Of genes; producing the same phenotype whether its allele is identical or dissimilar

Recessive:

Of genes; producing its characteristic phenotype only when its allele is identical

Incompletely dominant/recessive:

When heterozygous, giving an intermediate phenotype

Co-dominant:

When heterozygous, expressing both alleles

Sex-linked dominant/recessive:

Of genes; on the Z chromosome

Linkages:

Where loci do not segregate independently

Sex-Limited:

Where both genders carrying the gene, but gene expression is with one gender only (eg, egg-shell colour can only be expressed in hens)

Sex-Influenced:

Where a gene may appear dominant with one gender, recessive in another, ie gene expression is different between genders when alleles heterozygous

Autosomal:

Any chromosome that is not a sex chromosome; appear in pairs in body cells

Others:

Pleiotrophy:

Multiple traits expressed by a single gene

Polygenic trait:

Multiple genes giving accumulative effect on trait expression

Epistasis

Gene expression affected by a gene from another locus

Hypostasis:

The converse of epistasis, applied to the gene pair hidden by the epistatic gene pair

Penetrance:

The proportion of individuals of a specified genotype that express the expected phenotype

Expressivity:

The range of phenotypes expressed by a given genotype

Homologous Chromosomes:

Corresponding or similar in position

Multiple Alleles :

Where more than one mutation has occurred on a specific locus

Diploid:

An organism or cell having two sets of chromosomes

Pheomelanin:

Pigments that account for wild type red colour

Eumelanin:

Pigments that account for wild type black colour

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Genetics Theory:

Chickens have 78 chromosomes. They are diploid animals, therefore the body cell chromosomes are grouped together in pairs- 39. For example, a chicken will have two Chromosome 1’s, two chromosome 2’s, two chromosome 3’s, etc. The exception is the sex-chromosomes, Z and W, where roosters have two Z chromosomes and hens have only one Z chromosome, plus one W chromosome.

The following diagram may be useful in understanding the genetics terms:

The position along a chromosome where a gene may be found, is its locus. “Alleles” are the alternate genes possible on a specific locus. For example, the lavender gene (lav) may only be found on the Lavender locus. As only one mutation has been known to occur at this locus, there is a total of two alleles possible, ie the lav mutation and the wild type allele – Lav+ (wild type alleles are identified by the plus symbol: + ). Therefore, the possible locus allele combinations on the two homologous chromosomes are:

– Lav+ / Lav+

– Lav+ / lav

– lav / Lav+

– lav / lav

Sometimes multiple alleles have mutated on a given locus. That is, there may be a list of alleles available, not just two (ie, not just one mutation & the wild type allele). For example, E, ER, ER– Fayoumi, eWh, e+, eb, es, ebc, ey, eq alleles are all on the one locus (E locus). But only two alleles are physically possible in a diploid individual, at any given locus, ie, an individual may have ER/e+, or E/ER, or eb /E, etc, but only two alleles from the E Series.

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As mentioned previously, sex chromosomes are different to the above autosome pairs, and different again between roosters (2 Z chromosomes) and hens (W & Z chromosomes). The following diagrams may be useful in understanding the differences between male & female sex chromosome pairs:

* Z chromosome pair- Males only

* Z & W chromosome pair- Females only

Males may have two alleles the same- homozygous, or if two different alleles- heterozygous (two different alleles on the one locus), but as females can only have one allele, as only one Z chromosome present, they are referred to as hemizygous.

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Wild type (Red Jungle Fowl – Gallus gallus)

The “Wild Type” is what the wild species population appears like. For the domestic fowl, Jaap and Hollander (1954) suggested that the wild type Red Jungle Fowl (Gallus gallus) be taken as the standard from which all deviations (or mutations) are measured (Carefoot, 1985).

Pheomelanin and Eumelanin Pigment Areas

“Pheomelanin” equals the red pigment areas, “Eumelanin” equals the black pigment areas. Some genes affect the pheomelanin areas only, eg., if a bird has the cream- ig gene, the red is diluted to cream, if silver- S mutation present, the pheomelanin red areas are changed to silver -white, if mahogany- Mh, the red is darkened to deep mahogany, etc. Some genes affect the eumelanin areas only, eg., if a bird has the blue, dominant white, dun, smoky, etc genes, the eumelanin is diluted to these colours. Other genes affect both eumelanin and pheomelanin, eg the Lavender- lav gene dilutes the eumelanin- black to a pale lavender shade, and the pheomelanin- red to a pale straw-cream shade. The Recessive White gene changes all pigmented plumage (pheomelanin and eumelanin) to white, regardless of other mutations in the genotype.

Inheritance Modes, using Punnett Squares:

* Images used in the following punnet squares are from Feathersite, and the OEGBCA website.

Dominant / Recessive Inheritance

This inheritance mode is where one dose only of a dominant gene is needed for expression to occur (ie heterozygotes & homozygotes), but two doses are needed for expression of a recessive gene (ie homozygotes only express the gene). The example given below is of Lavender (lav- recessive gene) bird paired with a Black (Lav+ – dominant – wild type allele) bird.

Incomplete Dominant Inheritance

In the dominant/ recessive example, the Black, carrying lavender (heterozygote) bird is not distinguishable from the homozygous Black (not carrying lavender gene), as the black (wild type- Lav+) gene is completely dominant to the lavender gene. With Incompletely Dominant inheritance, the heterozygote gives an intermediate phenotype. For example, a splash (whitish- pale blue) bird (two doses of the Blue gene- Bl/Bl) paired to a black bird (no blue genes – bl+/bl+ -wild type) will produce an intermediate shade between the whitish/splash and black, ie- blue. The following is a punnet square example of this splash to black cross:

Sex-Linked Inheritance

Sex-Linked genes are genes on the Z chromosome. Roosters have two Z chromosomes and hens have only one Z chromosome, plus one W chromosome. As the hens only have one sex-linked allele for any given locus, both dominant and recessive sex-linked genes are expressed with just one gene (as hemizygous).

If the sex-linked recessive allele is homozygous in the male, and the sex-linked dominant allele is found in the female, all offspring males will inherit the sex-linked dominant gene from the mother, and all offspring females will inherit the sex-linked recessive gene from the fathers. The following is a punnet square example of a sex-linked cross, where all offspring females are the colour of the father and all offspring males are the colour of the mother. The recipricol cross (ie parent males with sex-linked dominant genes, parent females with sex-linked recessive gene) doesn’t work the same, as all offspring will have the dominant gene.

Dihybrid and other Multiple Mutation Crosses

When more than one mutation is in the equation, punnett squares become cumbersome. This is where the “Product Rule” comes in to play.

Product Rule:
prob(a and b) = p(a)p(b)

*which, in simple terms, is determined by finding the probability of each segregation that may occur in a pairing, then multiplying the probabilities together. For example, if it is determined that 1/4 of offspring will be Lavender, and 1/4 of offspring will be Mottled, it’s a simple matter of multiplying the probabilities together to find the probability of segregating both Lavender and Mottled in an individual:

Linkages

– Linkages occur where loci do not assort independently. Generally loci do assort independently, with crossover events common (eg, the recessive white gene assorts independently from the Pea Comb gene, as no linkage between their loci). For there to be a linkage between loci, they must be on the same homologous chromosome and in close proximity (less than 50 map units). Linked loci crossover less than 50 percent, as once they reach this distance, they assort independently. It is not uncommon for loci to assort independently of others even if on the same chromosome, as chromosomes can be very long and obtain many loci. For example, the S- Silver locus is not linked with the B – Barring locus, as over 50 map units apart on the Z chromosome.

The distance between the loci is measured in map units (centiMorgans) and is determined by the following equation:

Number of Recombinants

100 X

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= % of recombinants

Number of Offspring

* 1 map unit, or 1 centiMorgan is equal to 1% recombination

* The smaller the percentage of recombinants, the smaller the map distance between two loci.

A real example of close linkage is between the P (pea comb) locus from the O (Blue egg-shell), approximately 5 map units apart. Therefore, recombinants occur 5 % of the time. This is determined by crossing P-O P-O (pea-combed, blue egg-shell) X p+-o+ p+-o+ (single-combed, non-blue egg-shell) (or the opposite, crossing P-o+ P-o+ with p+-O p+-O), then breeding the F1 dihybrid offspring together. “Recombinants” is the total of both crossover segregates, ie both P/P o+/o+ (P-o+ P-o+) pea combed, non-blue egg-shell and p+/p+ O/O (p+-O p+-O) single combed, blue egg-shell) offspring (or with the second pairing, segregating F2 P-O P-O & p+-o+ p+-o+, as crossovers can occur both ways). So the number of single-combed blue egg-shell recombinants is half, ie 2.5% (2.5 in every 100 F2 offspring).

Sex-Linked, Linked Loci
It is also possible for linkages to occur on the Z chromosome, ie both sex-linked & linked loci. As only males in avian species have a homologous set of sex chromosomes (ie, only males have two Z chromosomes), crossovers can only occur with males.
For example, if the mo & lav loci were on the Z chromosome (ie sex-linked), with a 10% linkage, the following results would occur:

* Note, the above is an example only. The lav locus is NOT linked to the mo locus, & both are autosomes, NOT sex-linked.

Once again, the crossover frequency for any two genes is determined from breeding results, but the fact that crossovers only occur with the males (with 2 Z chromosomes), this needs to be taken into consideration with the calculations. The daughters obtain their Z chromosome from the fathers, the reason why the crossovers are expressed in the females, with the above example. In the above example, since black mottled & solid lavender female offspring can only be produced if crossover occurs, you would record the number of this type offspring produced, divide this number by the total number of offspring, times by 100 (for percentage), then multiply this number by 2 to determine the total crossover rate in percentage. The female recombinant percentage is multplied by two, to obtain the total recombinant rate for both males & females:

Epistasis Dihybrid Inheritance Modes

Epistasis is where gene expression is affected by a gene from another locus. As noted above, usually the F2 offspring from a dihybrid pairing (genes not linked) segregate to the phenotypic ratio of 9:3:3:1. Where there is epistasis, this phenotypic ratio changes, eg:

Dominant Epistasis: 12:3:1

Recessive Epistasis: 9:3:4

* Therefore, the dihybrid ratios can be used to determine whether epistasis is affecting inheritance.

Dominant Epistatis Inheritance:

The following diagram is of Dominant Epistatis inheritance, ie E is epistatic – & masks expression of Co (hypostatic). The Columbian gene – Co (on its own – ie no Db, etc) is not expressed on the E, ER alleles (regardless that Co is a dominant gene), but Co expresses the Columbian phenotype on eWh, e+, eb, ey, etc alleles.

* Note the above is not taking into consideration eumelanisers, nor incomplete dominance of E to eWh, etc, ie, the “black” phenotype in some may actually be closer to Brown Red/Grey phenotypes, etc depending on what other modifiers are present.

Recessive Epistatis Inheritance:

The following diagram is of Recessive Epistatis inheritance, ie Recessive white (c) is epistatic -masks all other plumage colours when homozygous, although a recessive gene:

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Variable Expressivity & Reduced Penetrance

These two are totally different concepts, with different statistical results. “Variable Expression” complies with expected inheritance ratios, & only relates to variance of phenotype in a bird, compared to his/her mates with the same genotype (eg light gold to dark gold variance, but all (100%) of birds with the same genotype expressing some shade of gold). “Penetrance” refers to how many (%) of the population that express the expected phenotype of a specific genotype, ie more to do with probabilities & the fact that some mutations don’t always express (eg some birds of the population with gold gene, not expressing any gold at all). Some mutations/genotypes have both variable expression & reduced penetrance.
Hopefully the following definitions & diagram will be clearer:

Definitions from “Genetics and Evolution of the Domestic Fowl” (by Lewis Stevens, 1991).

quote:

Expressivity:
The range of phenotypes expressed by a given genotype under any given set of environmental conditions.Penetrance:
The proportion of individuals of a specified genotype that show the expected phenotype under a defined set of environmental conditions.